Let $f$ be a function defined for all real numbers except for $1$. Also let $f'$, the derivative of $f$, be defined as $f'(x)=-\dfrac{(x-2)^2}{(x-1)^2}$. On which intervals is $f$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $1<x<2$ and $x>2$ (Choice B) B $x<1$ and $1<x<2$ (Choice C) C $x>2$ only (Choice D) D $x<1$ only (Choice E) E The entire domain of $f$
Explanation: We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $f'(x)=-\dfrac{(x-2)^2}{(x-1)^2}$ and that $f$ is undefined at $x=1$. $f'(x)=0$ for $x=2$. Our critical point is $x=2$, and we should also consider $x=1$. Our points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $ x<1$ $ 1<x<2$ $ x>2$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<1$ $x=-2$ $f'(-2)=-\dfrac{16}{9}<0$ $f$ is decreasing $\searrow$ $1<x<2$ $x=\dfrac{3}{2}$ $f'\left(\dfrac32\right)=-1<0$ $f$ is decreasing $\searrow$ $x>2$ $x=3$ $f'(3)=-\dfrac14<0$ $f$ is decreasing $\searrow$ In conclusion, $f$ is decreasing over its entire domain.